Functional Analysis/Hilbert spaces

(June 4, 2008) - The chapter is almost done, but there are still some errors in the proofs that have to be rectified. (Also, we could add a discussion of the polar decomposition of unbounded operators.)

A normed space is called a pre-Hilbert space if for each pair (x,y){\displaystyle (x,y)} of elements in the space there is a unique complex (or real) number called an inner product of x{\displaystyle x} and y{\displaystyle y}, denoted by ⟨x,y⟩{\displaystyle \langle x,y\rangle }, subject to the following conditions:

The inner product in its second variable is not linear but antilinear: i.e., if g(y)=⟨x,y⟩{\displaystyle g(y)=\langle x,y\rangle }, then g(αy)=α¯y{\displaystyle g(\alpha y)={\bar {\alpha }}y} for scalars α{\displaystyle \alpha }. We define ‖x‖=⟨x,x⟩1/2{\displaystyle \|x\|=\langle x,x\rangle ^{1/2}} and this becomes a norm. Indeed, it is clear that ‖αx‖=|α|‖x‖{\displaystyle \|\alpha x\|=|\alpha |\|x\|} and (iii) is the reason that ‖x‖=0{\displaystyle \|x\|=0} implies that x=0{\displaystyle x=0}. Finally, the triangular inequality follows from the next lemma.

3.1 Lemma (Schwarz's inequality)|⟨x,y⟩|≤‖x‖‖y‖{\displaystyle |\langle x,y\rangle |\leq \|x\|\|y\|} where the equality holds if and only if we can write x=λy{\displaystyle x=\lambda y} for some scalar λ{\displaystyle \lambda }.

where the equation becomes 0{\displaystyle 0} if and only if x=λy{\displaystyle x=\lambda y}. Since we may suppose that x≠0{\displaystyle x\neq 0}, the general case follows easily. ◻{\displaystyle \square }

3.2 TheoremA normed linear space is a pre-Hilbert space if and only if ‖x−y‖2=2‖x‖2+2‖y‖2−‖x+y‖2{\displaystyle \|x-y\|^{2}=2\|x\|^{2}+2\|y\|^{2}-\|x+y\|^{2}}.
Proof: The direct part is clear. To show the converse, we define

3.3 LemmaLet H{\displaystyle {\mathfrak {H}}} be a pre-Hilbert. Then xj→x{\displaystyle x_{j}\to x} in norm if and only if for any y∈H{\displaystyle y\in {\mathfrak {H}}}‖xj‖→‖x‖{\displaystyle \|x_{j}\|\to \|x\|} and ⟨xj−x,y⟩→0{\displaystyle \langle x_{j}-x,y\rangle \to 0} as j→∞{\displaystyle j\to \infty }. Proof: The direct part holds since:

This is to say, xn{\displaystyle x_{n}} is Cauchy. Since D{\displaystyle D} is a closed subset of a complete metric space, whence it is complete, there is a limit z∈D{\displaystyle z\in D} with ‖z‖=δ{\displaystyle \|z\|=\delta }. The uniqueness follows since if ‖w‖=δ{\displaystyle \|w\|=\delta } we have

where the right side is ≤0{\displaystyle \leq 0} for the same reason as before. ◻{\displaystyle \square }

The lemma may hold for a certain Banach space that is not a Hilbert space; this question will be investigated in the next chapter.

For a nonempty subset E⊂H{\displaystyle E\subset {\mathfrak {H}}}, define E⊥{\displaystyle E^{\bot }} to be the intersection of the kernel of the linear functional u↦⟨u,v⟩{\displaystyle u\mapsto \langle u,v\rangle } taken all over v∈E{\displaystyle v\in E}. (In other words, E⊥{\displaystyle E^{\bot }} is the set of all x∈H{\displaystyle x\in {\mathfrak {H}}} that is orthogonal to every y∈E{\displaystyle y\in E}.) Since the kernel of a continuous function is closed and the intersection of linear spaces is again a linear space, E⊥{\displaystyle E^{\bot }} is a closed (linear) subspace of H{\displaystyle {\mathfrak {H}}}. Finally, if x∈E∩E⊥{\displaystyle x\in E\cap E^{\bot }}, then 0=⟨x,x⟩=‖x‖{\displaystyle 0=\langle x,x\rangle =\|x\|} and x=0{\displaystyle x=0}.

3.5 LemmaLet M{\displaystyle {\mathcal {M}}} be a linear subspace of a pre-Hilbert space. Then z∈M⊥{\displaystyle z\in {\mathcal {M}}^{\bot }} if and only if ‖z‖=inf{‖z+w‖;w∈M}{\displaystyle \|z\|=\inf\{\|z+w\|;w\in {\mathcal {M}}\}}. Proof: The Schwarz inequality says the inequality

is actually equality if and only if z{\displaystyle z} and z+w{\displaystyle z+w} are linear dependent. ◻{\displaystyle \square } (TODO: the proof isn't quite well written.)

3.6 Theorem (orthogonal decomposition)Let H{\displaystyle {\mathfrak {H}}} be a Hilbert space and M⊂H{\displaystyle {\mathcal {M}}\subset {\mathfrak {H}}} be a closed subspace. For every x∈H{\displaystyle x\in {\mathfrak {H}}} we can write

x=y+z{\displaystyle x=y+z}

where y∈M{\displaystyle y\in {\mathcal {M}}} and z∈M⊥{\displaystyle z\in {\mathcal {M}}^{\bot }}, and y{\displaystyle y} and z{\displaystyle z} are uniquely determined by x{\displaystyle x}. Proof: Clearly x−M{\displaystyle x-{\mathcal {M}}} is convex, and it is also closed since a translation of closed set is again closed. Lemma 3.4 now gives a unique element y∈M{\displaystyle y\in {\mathcal {M}}} such that ‖x−y‖=inf{‖x−w‖;w∈M}{\displaystyle \|x-y\|=\inf\{\|x-w\|;w\in {\mathcal {M}}\}}. Let z=x−y{\displaystyle z=x-y}. By Lemma 3.5, z∈M⊥{\displaystyle z\in {\mathcal {M}}^{\bot }}. For the uniqueness, suppose we have written:

Proof: Let M=f−1({0}){\displaystyle {\mathcal {M}}=f^{-1}(\{0\})}. Since f{\displaystyle f} is continuous, M{\displaystyle {\mathcal {M}}} is closed. If M=H{\displaystyle {\mathcal {M}}={\mathfrak {H}}}, then take y=0{\displaystyle y=0}. If not, by Corollary 3.6, there is a nonzero z∈H{\displaystyle z\in {\mathfrak {H}}} orthogonal to M{\displaystyle {\mathcal {M}}}. By replacing z{\displaystyle z} with z‖z‖−1{\displaystyle z\|z\|^{-1}} we may suppose that ‖z‖=1{\displaystyle \|z\|=1}. For any x∈H{\displaystyle x\in {\mathfrak {H}}}, since zf(x)−f(z)x{\displaystyle zf(x)-f(z)x} is in the kernel of f{\displaystyle f} and thus is orthogonal to z{\displaystyle z}, we have:

In view of Theorem 3.5, for each x∈H{\displaystyle x\in {\mathfrak {H}}}, we can write: x=y+z{\displaystyle x=y+z} where y∈M{\displaystyle y\in {\mathcal {M}}}, a closed subspace of H{\displaystyle {\mathfrak {H}}}, and z∈M⊥{\displaystyle z\in {\mathcal {M}}^{\bot }}. Denote each y{\displaystyle y}, which is uniquely determined by x{\displaystyle x}, by π(x){\displaystyle \pi (x)}. The function π{\displaystyle \pi } then turns out to be a linear operator. Indeed, for given x1,x2∈H{\displaystyle x_{1},x_{2}\in {\mathfrak {H}}}, we write:

That is, π{\displaystyle \pi } is continuous with ‖π‖≤1{\displaystyle \|\pi \|\leq 1}. In particular, when M{\displaystyle {\mathcal {M}}} is a nonzero space, there is x0∈M{\displaystyle x_{0}\in {\mathcal {M}}} with π(x0)=x0{\displaystyle \pi (x_{0})=x_{0}} and ‖x0‖=1{\displaystyle \|x_{0}\|=1} and consequently ‖π‖=1{\displaystyle \|\pi \|=1}. Such π{\displaystyle \pi } is called an orthogonal projection (onto M{\displaystyle {\mathcal {M}}}).

The next theorem gives an alternative proof of the Hahn-Banach theorem.

3 TheoremLet M{\displaystyle {\mathcal {M}}} be a linear (not necessarily closed) subspace of a Hilbert space. Every continuous linear functional on M{\displaystyle {\mathcal {M}}} can be extended to a unique continuous linear functional on H{\displaystyle {\mathfrak {H}}} that has the same norm and vanishes on M⊥{\displaystyle {\mathcal {M}}^{\bot }}.
Proof: Since M{\displaystyle {\mathcal {M}}} is a dense subset of a Banach space M¯{\displaystyle {\overline {\mathcal {M}}}}, by Theorem 2.something, we can uniquely extend f{\displaystyle f} so that it is continuous on M¯{\displaystyle {\overline {\mathcal {M}}}}. Define g=f∘πM¯{\displaystyle g=f\circ \pi _{\overline {\mathcal {M}}}}. By the same argument used in the proof of Theorem 2.something (Hahn-Banach) and the fact that ‖πF‖=1{\displaystyle \|\pi _{\mathcal {F}}\|=1}, we obtain ‖f‖=‖g‖{\displaystyle \|f\|=\|g\|}. Since g=0{\displaystyle g=0} on M⊥{\displaystyle {\mathcal {M}}^{\bot }}, it remains to show the uniqueness. For this, let h{\displaystyle h} be another extension with the desired properties. Since the kernel of f−h{\displaystyle f-h} is closed and thus contain M¯{\displaystyle {\overline {\mathcal {M}}}}, f=h{\displaystyle f=h} on M¯{\displaystyle {\overline {\mathcal {M}}}}. Hence, for any x∈H{\displaystyle x\in {\mathfrak {H}}},

It is then easy to verify that (H1⊕H2,⟨⋅,⋅⟩){\displaystyle ({\mathfrak {H}}_{1}\oplus {\mathfrak {H}}_{2},\langle \cdot ,\cdot \rangle )} is a Hilbert space. It is also clear that this definition generalizes to a finite direct sum of Hilbert spaces. (For an infinite direct sum of Hilbert spaces, see Chapter 5.)

Recall from the previous chapter that an isometric surjection between Banach spaces is called "unitary".

That is to say, 0⊕u∈M{\displaystyle 0\oplus u\in {\mathcal {M}}}, which is a graph of a linear operator by assumption. Thus, u=0{\displaystyle u=0}. For the converse, suppose f⊕u1,f⊕u2∈M{\displaystyle f\oplus u_{1},f\oplus u_{2}\in {\mathcal {M}}}. Then

and so ⟨u1−u2,v⟩1=0{\displaystyle \langle u_{1}-u_{2},v\rangle _{1}=0} for every v{\displaystyle v} in the domain of T{\displaystyle T}, dense. Thus, u1=u2{\displaystyle u_{1}=u_{2}}, and M{\displaystyle {\mathcal {M}}} is a graph of a function, say, S{\displaystyle S}. The linear of S{\displaystyle S} can be checked in the similar manner.◻{\displaystyle \square }

Remark: In the proof of the lemma, the linear of T{\displaystyle T} was never used.

For a densely defined T{\displaystyle T}, we thus obtained a linear operator which we call T∗{\displaystyle T^{*}}. It is characterized uniquely by:

is continuous for every u∈dom⁡T{\displaystyle u\in \operatorname {dom} T}. The operator T∗{\displaystyle T^{*}} is called the Hilbert adjoint (or just adjoint) of T{\displaystyle T}. If T{\displaystyle T} is closed in addition to having dense domain, then

Here, the left-hand side is a graph of T∗∗{\displaystyle T^{**}}. For the second identity, since gra⁡S{\displaystyle \operatorname {gra} S} is a Hilbert space, it suffices to show gra⁡T⊥∩gra⁡S={0}{\displaystyle \operatorname {gra} T^{\bot }\cap \operatorname {gra} S=\{0\}}. But this follows from Lemma 3.something.◻{\displaystyle \square }

The next corollary is obvious but is important in application.

3 CorollaryLet H1,H2{\displaystyle {\mathfrak {H}}_{1},{\mathfrak {H}}_{2}} be Hilbert spaces, and T:H1→H2{\displaystyle T:{\mathfrak {H}}_{1}\to {\mathfrak {H}}_{2}} a closed densely defined linear operator. Then u∈dom⁡T{\displaystyle u\in \operatorname {dom} T} if and only if there is some K>0{\displaystyle K>0} such that:

3 LemmaLet T:H1→H2{\displaystyle T:{\mathfrak {H}}_{1}\to {\mathfrak {H}}_{2}} be a densely defined linear operator. Then ker⁡T∗=(ran⁡T)⊥.{\displaystyle \operatorname {ker} T^{*}=(\operatorname {ran} T)^{\bot }.}
Proof: f{\displaystyle f} is in either the left-hand side or the right-hand side if and only if:

In particular, a closed densely defined operator has closed kernel. As an application we shall prove the next theorem.

3 TheoremLet T:H1→H2{\displaystyle T:{\mathfrak {H}}_{1}\to {\mathfrak {H}}_{2}} be a closed densely defined linear operator. Then T{\displaystyle T} is surjective if and only if there is a K>0{\displaystyle K>0} such that

The last inequality holds since G{\displaystyle G} is continuous by the closed graph theorem. To show the converse, let g∈H2{\displaystyle g\in {\mathfrak {H}}_{2}} be given. Since T∗{\displaystyle T^{*}} is injective, we can define a linear functional L{\displaystyle L} by L(T∗f)=⟨f,g⟩2{\displaystyle L(T^{*}f)=\langle f,g\rangle _{2}} for f∈H2{\displaystyle f\in {\mathfrak {H}}_{2}}.,

Thus, L{\displaystyle L} is continuous on the range of T∗{\displaystyle T^{*}}. It follows from the Hahn-Banach theorem that we may assume that L{\displaystyle L} is defined and continuous on H1{\displaystyle {\mathfrak {H}}_{1}}. Thus, by Theorem 3.something, we can write L(⋅)=⟨⋅,u⟩1{\displaystyle L(\cdot )=\langle \cdot ,u\rangle _{1}} in H1{\displaystyle {\mathfrak {H}}_{1}} with some u{\displaystyle u}. Since L(T∗f){\displaystyle L(T^{*}f)} is continuous for f∈dom⁡T∗{\displaystyle f\in \operatorname {dom} T^{*}},

Thus, fj⊕S∗fj{\displaystyle f_{j}\oplus S^{*}f_{j}} is Cauchy in the graph of S∗{\displaystyle S^{*}}, which is closed. Hence, S∗fj{\displaystyle S^{*}f_{j}} converges within the range of S∗{\displaystyle S^{*}}. The converse holds since T∗∗=T{\displaystyle T^{**}=T}. ◻{\displaystyle \square }

We shall now consider some concrete examples of densely defined linear operators.

3 TheoremT:H1→H2{\displaystyle T:{\mathfrak {H}}_{1}\to {\mathfrak {H}}_{2}} is continuous if and only if T∗{\displaystyle T^{*}} is continuous. Moreover, when T{\displaystyle T} is continuous,

Proof: It is clear that T∗{\displaystyle T^{*}} is defined everywhere, and its continuity is a consequence of the closed graph theorem. Conversely, if T∗{\displaystyle T^{*}} is continuous, then T∗∗{\displaystyle T^{**}} is continuous and T=T∗∗{\displaystyle T=T^{**}}. For the second part,

Remark: the above lemma is false if the underlying field is R{\displaystyle \mathbf {R} }.

Recall that an isometric surjection is called unitary.

3 CorollaryA linear operator U:H1→H2{\displaystyle U:{\mathfrak {H}}_{1}\to {\mathfrak {H}}_{2}} is unitary if and only if U∗U{\displaystyle U^{*}U} and UU∗{\displaystyle UU^{*}} are identities.
Proof: Since (U∗Ux∣x)=‖Ux‖2=(x∣x)>{\displaystyle (U^{*}Ux\mid x)=\|Ux\|^{2}=(x\mid x)>}, we see that U∗U{\displaystyle U^{*}U} is the identity. Since UU∗U=U{\displaystyle UU^{*}U=U}, UU∗{\displaystyle UU^{*}} is the identity on the range of U, which is H2{\displaystyle {\mathcal {H}}_{2}} by surjectivity. Conversely, since ‖Ux‖22=⟨U∗Ux,x⟩1=‖x‖12{\displaystyle \|Ux\|_{2}^{2}=\langle U^{*}Ux,x\rangle _{1}=\|x\|_{1}^{2}}, U{\displaystyle U} is an isometry. ◻{\displaystyle \square }

Curiously, the hypothesis on linearity can be omitted:

3 TheoremIf U:H1→H2{\displaystyle U:{\mathcal {H}}_{1}\to {\mathcal {H}}_{2}} is a function such that

‖U(x)−U(y)‖2=‖x−y‖1{\displaystyle \|U(x)-U(y)\|_{2}=\|x-y\|_{1}}

for every x and y and U(0)=0{\displaystyle U(0)=0}, then U{\displaystyle U} is a linear operator (and so unitary). Proof: Note that U is continuous. Since ‖U(x)‖=‖U(x)−U(0)‖=‖x‖{\displaystyle \|U(x)\|=\|U(x)-U(0)\|=\|x\|}, we have:

3 ExerciseConstruct an example so as to show that an isometric operator (i.e., a linear operator that preserves norm) need not be unitary. (Hint: a shift operator.)

A densely defined linear operator T{\displaystyle T} is called "symmetric" if gra⁡T⊂gra⁡T∗{\displaystyle \operatorname {gra} T\subset \operatorname {gra} T^{*}}. If the equality in the above holds, then T{\displaystyle T} is called "self-adjoint". In light of Theorem 3.something, every self-adjoint is closed and densely defined. If T{\displaystyle T} is symmetric, then since T∗∗{\displaystyle T^{**}} is an extension of T{\displaystyle T},

But, by definition, (T2∘T1)∗u{\displaystyle (T_{2}\circ T_{1})^{*}u} denotes T1∗∘T2∗u{\displaystyle T_{1}^{*}\circ T_{2}^{*}u}. Hence, (T2∘T1)∗{\displaystyle (T_{2}\circ T_{1})^{*}} is an extension of T1∗∘T2∗{\displaystyle T_{1}^{*}\circ T_{2}^{*}}. For the second part, the fact we have just proved gives:

3 TheoremLet T:H1→H2{\displaystyle T:{\mathfrak {H}}_{1}\to {\mathfrak {H}}_{2}} be a Hilbert spaces. If T:H1→H2{\displaystyle T:{\mathfrak {H}}_{1}\to {\mathfrak {H}}_{2}} is a closed densely defined operator, then T∗T{\displaystyle T^{*}T} is a self-adjoint operator (in particular, densely defined and closed.)
Proof: In light of the preceding theorem, it suffices to show that T∗T{\displaystyle T^{*}T} is closed. Let uj∈dom⁡T∗T{\displaystyle u_{j}\in \operatorname {dom} T^{*}T} be a sequence such that (uj,T∗Tuj){\displaystyle (u_{j},T^{*}Tu_{j})} converges to limit (u,v){\displaystyle (u,v)}. Since

there is some f∈H2{\displaystyle f\in {\mathfrak {H}}_{2}} such that: ‖Tuj−f‖2→0{\displaystyle \|Tu_{j}-f\|_{2}\to 0}. It follows from the closedness of T∗{\displaystyle T^{*}} that T∗f=v{\displaystyle T^{*}f=v}. Since ‖uj−u‖1+‖Tuj−f‖2→0{\displaystyle \|u_{j}-u\|_{1}+\|Tu_{j}-f\|_{2}\to 0} and T{\displaystyle T} is closed, T∗Tu=T∗f=v{\displaystyle T^{*}Tu=T^{*}f=v}. ◻{\displaystyle \square }

3 TheoremLet T{\displaystyle T} be a symmetric densely defined operator. If T{\displaystyle T} is surjective, then T{\displaystyle T} is self-adjoint and injective and T−1{\displaystyle T^{-1}} is self-adjoint and bounded.
Proof: If Tu=0{\displaystyle Tu=0},

Here, V(x1⊕x2)=x2⊕x1{\displaystyle V(x_{1}\oplus x_{2})=x_{2}\oplus x_{1}}, and the equality holds since the domains of T{\displaystyle T} and T∗{\displaystyle T^{*}} coincide. Hence, T−1{\displaystyle T^{-1}} is self-adjoint. Since we have just proved that the inverse of a self-adjoint is self-adjoint, we have: (T−1)−1{\displaystyle (T^{-1})^{-1}} is self-adjoint.◻{\displaystyle \square }

3 TheoremLet M{\displaystyle {\mathcal {M}}} be a closed linear subspace of a Hilbert space H{\displaystyle {\mathfrak {H}}}. Then π{\displaystyle \pi } is an orthogonal projection onto M{\displaystyle {\mathcal {M}}} if and only if π=π∗=π2{\displaystyle \pi =\pi ^{*}=\pi ^{2}} and the range of π{\displaystyle \pi } is M{\displaystyle {\mathcal {M}}}.
Proof: The direct part is clear except for π=π∗{\displaystyle \pi =\pi ^{*}}. But we have: